Ex 6.5 Class 7 Maths Question 1.

PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:

In right angled triangle PQR, we have

QR^{2} = PQ^{2} + PR^{2} From Pythagoras property)

= (10)^{2} + (24)^{2}

= 100 + 576 = 676

∴ QR = \(\sqrt{676}\) = 26 cm

The, the required length of QR = 26 cm.

Ex 6.5 Class 7 Maths Question 2.

ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Solution:

In right angled∆ABC, we have

BC^{2} + (7)^{2} = (25)^{2} (By Pythagoras property)

⇒ BC^{2} + 49 = 625

⇒ BC^{2} = 625 – 49

⇒ BC^{2} = 576

∴ BC = \(\sqrt{576}\) = 24 cm

Thus, the required length of BC = 24 cm.

Ex 6.5 Class 7 Maths Question 3.

A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:

Here, the ladder forms a right angled triangle.

∴ a^{2} + (12)^{2} = (15)^{2} (By Pythagoras property)

⇒ a^{2}+ 144 = 225

⇒ a2 = 225 – 144

⇒ a^{2} = 81

∴ a = \(\sqrt{81}\) = 9 m

Thus, the distance of the foot from the ladder = 9m

Ex 6.5 Class 7 Maths Question 4.

Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm.

(ii) 2 cm, 2 cm, 5 cm.

(iii) 1.5 cm, 2 cm, 2.5 cm

Solution:

(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.

Square of the longer side = (6.5)^{2} = 42.25 cm.

Sum of the square of other two sides

= (2.5)^{2} + (6)^{2} = 6.25 + 36

= 42.25 cm.

Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.

∴ The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .

Square of the longer side = (5)^{2} = 25 cm Sum of the square of other two sides

= (2)^{2} + (2)^{2} =4 + 4 = 8 cm

Since 25 cm ≠ 8 cm

∴ The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm

Square of the longer side = (2.5)^{2} = 6.25 cm Sum of the square of other two sides

= (1.5)^{2} + (2)^{2} = 2.25 + 4

Since 6.25 cm = 6.25 cm = 6.25 cm

Since the square of longer side in a triangle is equal to the sum of square of other two sides.

∴ The given sides form a right triangle.

Ex 6.5 Class 7 Maths Question 5.

A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.

Solution:

Let AB be the original height of the tree and broken at C touching the ground at D such that

AC = 5 m

and AD = 12 m

In right triangle ∆CAD,

AD^{2} + AC^{2} = CD^{2} (By Pythagoras property)

⇒ (12)^{2} + (5)^{2} = CD^{2}

⇒ 144 + 25 = CD^{2}

⇒ 169 = CD^{2}

∴ CD = \(\sqrt{169}\) = 13 m

But CD = BC

AC + CB = AB

5 m + 13 m = AB

∴ AB = 18 m .

Thus, the original height of the tree = 18 m.

Ex 6.5 Class 7 Maths Question 6.

Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.

(i) PQ^{2} + QR^{2} = RP^{2}

(ii) PQ^{2} + RP^{2} = QR^{2}

(iii) RP^{2} + QR^{2} = PQ^{2}

Solution:

We know that

∠P + ∠Q + ∠R = 180° (Angle sum property)

∠P + 25° + 65° = 180°

∠P + 90° = 180°

∠P = 180° – 90° – 90°

∆PQR is a right triangle, right angled at P

(i) Not True

∴ PQ^{2} + QR^{2} ≠ RP^{2} (By Pythagoras property)

(ii) True

∴ PQ^{2} + RP^{2} = QP^{2} (By Pythagoras property)

(iii) Not True

∴ RP^{2} + QR^{2} ≠ PQ^{2} (By Pythagoras property)

Ex 6.5 Class 7 Maths Question 7.

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Solution:

Given: Length AB = 40 cm

Diagonal AC = 41 cm

In right triangle ABC, we have

AB^{2} + BC^{2}= AC^{2} (By Pythagoras property)

⇒ (40)^{2} + BC^{2} = (41)^{2}

⇒ 1600 + BC^{2} = 1681

⇒ BC^{2} = 1681 – 1600

⇒ BC^{2} = 81

∴ BC = \(\sqrt{81}\) = 9 cm

∴ AB = DC = 40 cm and BC = AD = 9 cm (Property of rectangle)

∴ The required perimeter

= AB + BC + CD + DA

= (40 + 9 + 40 + 9) cm

= 98 cm

Ex 6.5 Class 7 Maths Question 8.

The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Solution:

Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm

Since, the diagonals of a rhombus bisect each other at 90°.

∴ OA = OC = 8 cm and OB = OD = 15 cm

In right ∆OAB,

AB^{2} = OA^{2} + OB^{2} (By Pythagoras property)

= (8)^{2}+ (15)^{2} = 64 + 225

= 289

∴ AB =\(\sqrt{289}\)= 17 cm

Since AB = BC = CD = DA (Property of rhombus)

∴ Required perimeter of rhombus

= 4 × side = 4 × 17 = 68 cm.